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Question no 18

18. Two identical calorimeters 'X' and 'Y', of water equivalent 10 g each, contain equal quantities of same liquid. The mass, initial temperature and specific heat capacity of the liquid present in both calorimeters are 50 g, 25°C and 12cal g-1°C-1 respectively. A 10 g metal piece (say A) of specific heat capacity 0.4 cal g-1°C-1is dropped in calorimeter "X" and 25 g of metal piece (say B) of specific heat capacity "SB" is dropped into calorimeter "Y". Due to this the equilibrium temperature in 'X' and 'Y' rises to 30°C and 40°C respectively. If the initial temperature of 'B' is twice of 'A', then find the value of "SB".

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Solution

Dear student


In calorimeter XHeat lost by metal A= heat gained by liquid calorimeter XmA.SA.(t)A={50g×0.5×(30-25)} +(5oC×10g)}10×0.4×(T-30oC)=(50×0.5×5+50 cal)--(1)In calorimeter YFind the initial temperature of metal B(2T).Heat lost by metal B=heat gained by (liquid+calorimeter)mB.SB(2T-40)=50g×0.5×(40-25)+(15oC×25g)Solving we getSB =0.2cal/g/oC
Regards

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